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Difference between revisions of "2013 AIME I Problems/Problem 14"

(→‎Solution)
(→‎Solution)
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===Solution 1===
 
===Solution 1===
 
<math>\begin{align*}</math>
 
<math>\begin{align*}</math>
−
<math>P sin\theta\ + Q cos\theta\ = cos\theta\ - \frac{1}{2}\ P</math>
+
<math>P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P</math>
 
<math>\end{align*}</math>
 
<math>\end{align*}</math>
 
and
 
and
 
<math>\begin{align*}</math>
 
<math>\begin{align*}</math>
−
<math>P cos\theta\ + Q sin\theta\ = -2(Q-1)</math>
+
<math>P \cos\theta\ + Q \sin\theta\ = -2(Q-1)</math>
 
<math>\end{align*}</math>
 
<math>\end{align*}</math>
  
Line 27: Line 27:
  
  
−
<math>\frac{P}{Q} = \frac{cos\theta\ ( sin\theta + 2)}{8 + 8sin\theta + 2sin^2\theta } = \frac{2\sqrt2}{7}</math>
+
<math>\frac{P}{Q} = \frac{\cos\theta\ ( \sin\theta + 2)}{8 + 8\sin\theta + 2\sin^2\theta } = \frac{2\sqrt2}{7}</math>
  
 
Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math>
 
Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math>
  
−
<math>sin\theta = -\frac{17}{19} </math>
+
<math>\sin\theta = -\frac{17}{19} </math>
  
 
The answer is <math>\boxed{036}</math>
 
The answer is <math>\boxed{036}</math>
Line 40: Line 40:
 
<math>\end{align*}</math>
 
<math>\end{align*}</math>
  
−
<math>P sin\theta\ + Q cos\theta\ = cos \theta\ - \frac{1}{4}\cos\theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math>
+
<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math>
  
 
<math>\begin{align*}</math>
 
<math>\begin{align*}</math>
−
Therefore, <math>P sin\theta\ - Q cos\theta\ = -2P</math>
+
Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math>
 
<math>\end{align*}</math>
 
<math>\end{align*}</math>
  

Revision as of 13:26, 24 March 2013

Problem 14

14. For $\pi \le \theta < 2\pi$, let

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

and

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg) and $\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P \cos\theta\ + Q \sin\theta\ = -2(Q-1)$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Solving for P, Q we have


$\frac{P}{Q} = \frac{\cos\theta\ ( \sin\theta + 2)}{8 + 8\sin\theta + 2\sin^2\theta } = \frac{2\sqrt2}{7}$

Square both side, and use polynomial rational root theorem to solve $sin\theta$

$\sin\theta = -\frac{17}{19}$

The answer is $\boxed{036}$

Solution 2

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Use sum to product formulas to rewrite $P$ and $Q$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...$

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Therefore, $P \sin \theta - Q \cos \theta = -2P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Using $\frac{P}{Q} = \frac{2\sqrt2}{7}$, $Q = \frac{7}{2\sqrt2} P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Plug in to the previous equation and cancel out the "P" terms to get: $sin\theta - \frac{7}{2\sqrt2} cos\theta = -2$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Then use the pythagorean identity to solve for $sin\theta$, $sin\theta = \boxed{036}$

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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